IB DP Mathematics: Analysis and Approaches – Functions Strand Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

Related: Functions study guide and revision notes.

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Section A

1. State the domain and range of $f(x) = \sqrt{x - 3}$. [2]

2. Given $f(x) = 2x - 1$, find $f^{-1}(x)$. [2]

3. Describe the transformation that maps $y = f(x)$ to $y = f(x) - 5$. [1]

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Section B

4. $f(x) = x^2 - 4$ and $g(x) = 3x + 1$.

(a) Find $f(g(1))$. [2] (b) Find $g(f(1))$. [2]

5. The graph of $y = x^2$ is transformed to $y = -3(x+2)^2 + 1$.

(a) Describe the sequence of transformations applied to $y = x^2$ to obtain this graph. [3] (b) State the coordinates of the resulting graph’s vertex, and whether it is a maximum or minimum point. [2]

6. Solve $2^{x+1} = 30$, giving your answer to 3 significant figures. [3]

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Section C

7. Solve $\log_2(x) + \log_2(x - 2) = 3$, showing why one of the two solutions to the resulting quadratic must be rejected. [5]

8. $f(x) = \dfrac{2x+1}{x-3}$, $x \neq 3$.

(a) Find $f^{-1}(x)$, and state the value $x$ must not take. [4] (b) Verify that $f^{-1}(f(0)) = 0$. [2]

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Worked answers

1. Domain: $x \geq 3$ (the expression under the root cannot be negative). Range: $f(x) \geq 0$ (a square root is never negative). [2]

2. Let $y = 2x - 1$, so $x = (y+1)/2$. Therefore $f^{-1}(x) = (x+1)/2$. [2]

3. A vertical translation of 5 units downward (a translation by $(0, -5)$). [1]

4. (a) $g(1) = 3(1) + 1 = 4$. $f(g(1)) = f(4) = 4^2 - 4 = 12$. [2] (b) $f(1) = 1^2 - 4 = -3$. $g(f(1)) = g(-3) = 3(-3) + 1 = -8$. [2]

5. (a) Starting from $y = x^2$: a horizontal translation of 2 units left, giving $y = (x+2)^2$; then a vertical stretch with scale factor 3 and a reflection in the $x$-axis, giving $y = -3(x+2)^2$; then a vertical translation of 1 unit up, giving $y = -3(x+2)^2 + 1$. [3] (b) The vertex of $y = x^2$ is $(0,0)$. The horizontal translation moves it to $(-2, 0)$; the stretch and reflection leave a vertex value of 0 unchanged (since $-3(0) = 0$), so it remains $(-2, 0)$; the vertical translation moves it to $(-2, 1)$. Since the coefficient of the squared term is negative ($-3$), this is a maximum point. [2]

6. Taking $\log$ of both sides: $(x+1)\ln 2 = \ln 30$, so $x + 1 = \dfrac{\ln 30}{\ln 2} = \dfrac{3.4012}{0.6931} = 4.9069$. Therefore $x = 3.9069… \approx 3.91$ (3 s.f.). [3]

7. Combining the logarithms: $\log_2[x(x-2)] = 3$, so $x(x-2) = 2^3 = 8$, giving $x^2 - 2x - 8 = 0$. Factorising: $(x-4)(x+2) = 0$, so $x = 4$ or $x = -2$. The original equation requires both $\log_2(x)$ and $\log_2(x-2)$ to be defined, which needs $x > 0$ and $x > 2$ simultaneously, i.e. $x > 2$. Since $x = -2$ does not satisfy this, it is rejected; only $x = 4$ is a valid solution. Checking: $\log_2(4) + \log_2(2) = 2 + 1 = 3$. [5]

8. (a) Let $y = \dfrac{2x+1}{x-3}$. Then $y(x-3) = 2x+1$, so $yx - 3y = 2x + 1$, giving $yx - 2x = 1 + 3y$, so $x(y-2) = 1 + 3y$, and $x = \dfrac{1+3y}{y-2}$. Therefore $f^{-1}(x) = \dfrac{1+3x}{x-2}$, and $x$ must not take the value $2$ (since this would make the denominator zero). [4] (b) $f(0) = \dfrac{2(0)+1}{0-3} = \dfrac{1}{-3} = -\dfrac{1}{3}$. Then $f^{-1}\left(-\dfrac{1}{3}\right) = \dfrac{1 + 3\left(-\frac{1}{3}\right)}{-\frac{1}{3} - 2} = \dfrac{1 - 1}{-\frac{7}{3}} = \dfrac{0}{-\frac{7}{3}} = 0$, confirming $f^{-1}(f(0)) = 0$. [2]

A note on method

Several of these questions (5, 7 and 8) are deliberately built so that a purely mechanical answer – applying a rule without checking it actually applies – gives a wrong or incomplete result. Question 5 rewards tracking what happens to the vertex specifically at each stage of a sequence of transformations, rather than only substituting a single coordinate into the final equation and hoping the transformations were applied in a sensible order. Question 7 is a reminder that solving a logarithmic equation by combining logs and clearing them algebraically produces a quadratic with two roots, but the logarithmic form itself restricts the domain, so checking each root against that restriction is not optional working – it is part of the method the mark scheme expects to see, and a numerically correct root that is not checked against the domain, or that is left in without comment, typically does not receive full credit. Question 8 similarly expects the excluded value of the inverse function’s domain to be stated explicitly, not left implicit, since a function defined by a fraction is only fully specified once its domain restriction is given alongside the algebraic rule. Building this habit – stating restrictions and justifying rejected solutions explicitly, rather than only writing down a final numerical answer – is worth practising across every Functions question, not just the ones where a restriction happens to be the main point being tested.

Official syllabus

International Baccalaureate Organization, Diploma Programme Subject Brief – Mathematics: Analysis and Approaches, first assessment 2021 – the same source cited by the study guide and revision notes. Verified 2026-09-06.